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sweet-ann [11.9K]
4 years ago
9

Olivia places her pet frog on a line to observe the frog motion. The line is divided into sections that measure 1 centimeter eac

h. the frog begins at 0, moves 18 centimeters forward, moves 6 centimeters backwards, and then 12 centimeters backward. What is the frog's displacement?
Physics
2 answers:
Goshia [24]4 years ago
4 0
There is no displacement. The frog is back where it began.
Natali [406]4 years ago
3 0

Answer:

displacement of frog is zero

Explanation:

Displacement of the frog is given as

d_1 = 18 cm forward

d_2 = 6 cm backwards

d_3 = 12 cm backwards

so here we have

d_1 = 18 cm

d_2 = -6 cm

d_3 = -12 cm

so we have

d = d_1 + d_2 + d_3

d = 18 - 6 - 12

d = 0

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gladu [14]

The work done by the heat engine is 40 kCal.

The given parameters;

  • input heat of the engine, Q₁ = 70 kCal
  • output heat of the engine, Q₂ = 30 kCal

To find:

  • the work done by the heat engine

The work done by the heat engine is the change in the heat energy of the engine;

W = Q₂ - Q₁

Substitute the given parameters and solve work done (W)

W = 70 kCal - 30 kal

W = 40 kCal

Thus, the work done by the heat engine is 40 kCal.

Learn more here: brainly.com/question/4280097

4 0
3 years ago
How much mass energy could be obtained from the complete conversion of a 235 g hamburger?
Radda [10]
E = mc²
E = 0.235 kg · (3×10⁸ m/s)² = 0.235 · 9×10¹⁶  kg·m/s²
E = 2.115×10¹⁶ J
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8 0
3 years ago
An object is moving with a constant velocity of 278 m/s. How long will it take it to travel 7500 m, using the formula Delta X=Vt
rusak2 [61]
If the velocity is constant then the acceleration of the object is zero.
a=0 (m/s^2)
Thus when we apply the equation
\Delta X=vt+(at^2/2)
It remains
\Delta X =vt
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t=(\Delta X/v) =7500/278 =26.98 (seconds)

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4 years ago
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Answer:

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Explanation:

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5 0
3 years ago
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What is the centripetal acceleration of the earth as it travels around the sun when Earth has an orbital radius of 1.5 x 10^11 m
masya89 [10]

Answer: 0.0058 m/s^{2}

Explanation:

Centripetal acceleration a_{C} is calculated by the following equation:

a_{C}=\frac{V^{2}}{r}

Where:

V=29.7 \frac{km}{h} \frac{1000 m}{1 km}=29700 m/s is the Earth's orbital speed

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a_{C}=0.0058 m/s^{2}

4 0
3 years ago
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